16. Integration of powers of the sine function (Wallis integral)

수학 모음 (Maths collection) 전체보기

 

Question. Let $$\begin{align} I_n = \int \sin^n(ax) \, \textrm{d}x \end{align}$$

(a) Show that $$\begin{align} I_n = -\frac1{an} \sin^{n-1}(ax) \cos(ax) + \frac{n-1}{n}I_{n-2} \end{align}$$

 

Let $$\begin{align} J_n = \int_0^{\frac{\pi}{2}} \sin^nx \, \textrm{d}x \end{align}$$

(b) Hence, or otherwise, show that $$\begin{align} J_n = \frac{n-1}{n}J_{n-2} \end{align}$$ and hence that $$\begin{align} \textrm{(i)}&& n &= 2m \; (\textrm{even}):& J_{2m}&=\int_0^{\frac{\pi}{2}}\sin^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \textrm{(ii)}&& n &= 2m+1 \; (\textrm{odd}):& J_{2m+1}&=\int_0^{\frac{\pi}{2}}\sin^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \end{align}$$

 

Comment. We can generalise this result to: $$\begin{align} K_{2m}&=\int_0^{\frac{\pi}{2a}}\sin^{2m}(ax)\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2a} \\ K_{2m+1}&=\int_0^{\frac{\pi}{2a}}\sin^{2m+1}(ax)\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!}\frac1a \end{align}$$

 

(Relevant: Edexcel 2017 Specification, FP2 Ch6 Reduction formulae)

 

 

Solution. (a) By integration by parts, $$\begin{align} u&=\sin^{n-1}(ax) & v'&=\sin(ax) \\ u'&=a(n-1)\sin^{n-2}(ax)\cos(ax)& v&=-\frac1a\cos(ax) \end{align}$$ we find $$\begin{align} && I_n &=\int\sin^n(ax)\,{\rm d}x \\ &&&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)\int \sin^{n-2}(ax)\cos^2(ax) \, {\rm d}x \\ &&&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)\int \sin^{n-2}(ax)\left(1-\sin^2(ax)\right)\,{\rm d}x \\ &&&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)\underbrace{\int \sin^{n-2}(ax)\,{\rm d}x}_{I_{n-2}}-(n-1)\underbrace{\int \sin^{n}(ax)\,{\rm d}x}_{I_{n}} \\ &&&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)I_{n-2}-(n-1)I_n \\ \\ &\Rightarrow& \Big[1+(n-1)\Big]I_n&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)I_{n-2} \\ &\Rightarrow& nI_n&=-\frac1a\sin^{n-1}(ax)\cos(ax) + (n-1)I_{n-2} \\ \\ &\Rightarrow& I_n&=-\frac1{an}\sin^{n-1}(ax)\cos(ax)+\frac{n-1}{n}I_{n-2} \qquad \square \end{align}$$

 

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(b) From (a) with $a=1$, $$\begin{align} J_n &=\int_0^{\frac{\pi}{2}}\sin^nx\,{\rm d}x \\ &=\underbrace{\left[-\frac1n\sin^{n-1}x\cos x\right]_0^{\frac{\pi}{2}}}_{=0}+\frac{n-1}{n}J_{n-2} \\ &=\frac{n-1}{n}J_{n-2} \qquad \square \end{align}$$

 

For the next part, we may approach it in three different ways.

 

Method 1.

We note the following results: $$\begin{align} J_0 &=\int_0^{\frac{\pi}{2}}\,{\rm d}x=\frac{\pi}{2} \\ J_1 &=\int_0^{\frac{\pi}{2}}\sin x\,{\rm d}x = \Big[-\cos x\Big]_0^{\frac{\pi}{2}} = -\cos\left(\frac{\pi}{2}\right)+\cos(0) = 1 \\ J_2 &=\int_0^{\frac{\pi}{2}}\sin^2 x\,{\rm d}x =\int_0^{\frac{\pi}{2}}\left(\frac{1-\cos 2x}{2}\right){\rm d}x = \left[\frac{1}{2}x-\frac14\sin2x\right]_0^{\frac{\pi}{2}} = \frac{\pi}{4} \left(=\frac12J_0\right) \end{align}$$ (i) $n=2m$ (even) $$\begin{align} J_{2m} &=\frac{2m-1}{2m}I_{2m-2} \\ &=\frac{2m-1}{2m}\cdot\frac{2m-3}{2m-2}J_{2m-4} \\ &=\frac{2m-1}{2m}\cdot\frac{2m-3}{2m-2}\cdot\frac{2m-5}{2m-4}J_{2m-6} \\ &=\frac{2m-1}{2m}\cdot\frac{2m-3}{2m-2}\cdot\frac{2m-5}{2m-4}\cdots\frac{3}{4}\cdot\frac12\underbrace{J_0}_{\frac{\pi}{2}} \\ &=\frac{(2m-1)(2m-3)(2m-5)\cdots3\cdot1}{2m(2m-2)(2m-4)\cdots4\cdot2}\cdot\frac{\pi}{2} \\ &=\frac{(2m)(2m-1)(2m-2)(2m-3)(2m-4)(2m-5)\cdots3\cdot2\cdot1}{\Big[2m(2m-2)(2m-4)\cdots4\cdot2\Big]^2}\frac{\pi}{2} \\ &=\frac{(2m)!}{\left[2^m\cdot m(m-1)(m-2)\cdots2\cdot1\right]^2}\frac{\pi}{2} \\ &=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \qquad \square \end{align}$$ (ii) $n=2m+1$ (odd) $$\begin{align} J_{2m+1} &=\frac{2m}{2m+1}J_{2m-1} \\ &=\frac{2m}{2m+1}\cdot\frac{2m-2}{2m-1}J_{2m-3} \\ &=\frac{2m}{2m+1}\cdot\frac{2m-2}{2m-1}\cdot\frac{2m-4}{2m-3}J_{2m-5} \\ &=\frac{2m}{2m+1}\cdot\frac{2m-2}{2m-1}\cdot\frac{2m-4}{2m-3}\cdots\frac{2}{1}\underbrace{J_1}_{=1} \\ &=\frac{2m(2m-2)(2m-4)\cdots4\cdot2}{(2m+1)(2m-1)(2m-1)\cdots3\cdot1} \\ &=\frac{\Big[2m(2m-2)(2m-4)\cdots4\cdot2\Big]^2}{(2m+1)(2m)(2m-1)(2m-2)(2m-3)(2m-4)\cdots3\cdot2\cdot1} \\ &=\frac{\left[2^m\cdot m(m-1)(m-2)\cdots2\cdot1\right]^2}{(2m+1)!} \\ &=\frac{2^{2m}(m!)^2}{(2m+1)!} \qquad \square \end{align}$$

 

Aside. As an exemplary application, we evaluate the first few integrals: $$\begin{align} J_0 &= \int_0^{\frac{\pi}{2}} \,{\rm d}x = \frac{ \pi }{ 2 } &&& J_1 &= \int_0^{\frac{\pi}{2}} \sin x \,{\rm d}x = 1 \\ J_2 &= \int_0^{\frac{\pi}{2}} \sin^2x \,{\rm d}x = \frac{ \pi }{ 4 } &&& J_3 &= \int_0^{\frac{\pi}{2}} \sin^3 x \,{\rm d}x = \frac{ 2 }{ 3 } \\ J_4 &= \int_0^{\frac{\pi}{2}} \sin^4x \,{\rm d}x = \frac{ 3 \pi }{ 16 } &&& J_5 &= \int_0^{\frac{\pi}{2}} \sin^5 x \,{\rm d}x = \frac{ 8 }{ 15 } \\ J_6 &= \int_0^{\frac{\pi}{2}} \sin^6x \,{\rm d}x = \frac{ 5 \pi }{ 32 } &&& J_7 &= \int_0^{\frac{\pi}{2}} \sin^7 x \,{\rm d}x = \frac{ 16 }{ 35 } \\ J_8 &= \int_0^{\frac{\pi}{2}} \sin^8x \,{\rm d}x = \frac{ 35 \pi }{ 256 } &&& J_9 &= \int_0^{\frac{\pi}{2}} \sin^9 x \,{\rm d}x = \frac{ 128 }{ 315 } \\ J_{10} &= \int_0^{\frac{\pi}{2}} \sin^{10}x \,{\rm d}x = \frac{ 63 \pi }{ 512 } &&& J_{11} &= \int_0^{\frac{\pi}{2}} \sin^{11} x \,{\rm d}x = \frac{ 256 }{ 693 } \end{align}$$

 

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Method 2. Exponential form and binomial theorem:

We can also use the exponential definition of the sine function, $$\begin{align} \sin x = \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \end{align}$$ (i) $n=2m$ (even) $$\begin{align} J_{2m} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m}} \left[ \sum_{r=0}^{2m} \binom{2m}{r} \textrm{e}^{i(2m-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align}$$

 

Here, we note that, for $r \ne m$, $$\begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{2i(m-r)x} \, \textrm{d}x \\ &= \left[ \frac1{2i(m-r)} \textrm{e}^{2i(m-r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac{1}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ \end{align}$$ which gives (ignoring the subtlety with $r=m$ for the next few lines as we're just trying to see the pattern for now) $$\begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} (-1)^r I_r \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \left( \textrm{e}^{i(m-r)\pi} - 1 \right) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r}{2i(m-r)} \Big( \cos[(m-r)\pi] - 1 \Big) \\ &= \frac1{(2i)^{2m}} \sum_{r=0}^{2m} \binom{2m}{r} \frac{(-1)^r(-1)}{i(m-r)} \sin^2 \left( \frac{(m-r)\pi}{2} \right) \end{align}$$ We observe a symmetry in the terms with $r=s$ and $r=2m-s$, $$\begin{align} r&=s \;:& a_s&=\binom{2m}{s} \frac{(-1)^s(-1)}{i(m-s)} \sin^2 \left( \frac{(m-s)\pi}{2} \right) \\ \\ r&=2m-s \;:& a_{2m-s}&=\binom{2m}{2m-s} \frac{(-1)^{2m-s}(-1)}{i(s-m)} \sin^2 \left( \frac{(s-m)\pi}{2} \right) \end{align}$$ and, since $\binom{2m}{s}=\binom{2m}{2m-s}$, we find $$\begin{align} a_s = - a_{2m-s} \end{align}$$ This motivates to separate the sum into three parts, $$\begin{align} \sum_{r=0}^{2m} = \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \end{align}$$ and in fact we should apply this separation before we perform the integral of $I_r$ and include the case $r=m$ as well: $$\begin{align} \Rightarrow \qquad J_{2m} &= \frac1{(2i)^{2m}} \left( \sum_{r=0}^{m-1} + \sum_{r=m}^{m} + \sum_{r=m+1}^{2m} \right) \binom{2m}{r} (-1)^r I_r \\ &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{r=m+1}^{2m} \binom{2m}{r} (-1)^r I_r \right] \end{align}$$ For the last summation, let $k = 2m-r$ $$\begin{align} \Rightarrow \qquad J_{2m} &= \frac{(-1)^m}{2^{2m}} \left[ \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r I_r + \binom{2m}{m} (-1)^m I_m + \sum_{k=0}^{m-1} \binom{2m}{2m-k} (-1)^{2m-k} I_{2m-k} \right] \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \Big( I_r + I_{2m-r} \Big) + \frac{(-1)^m}{2^{2m}} \binom{2m}{m} (-1)^m I_m \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \left( \frac{\textrm{e}^{i(m-r)\pi} - 1}{2i(m-r)} + \frac{\textrm{e}^{-i(m-r)\pi} - 1}{-2i(m-r)} \right) + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m-1} \binom{2m}{r} (-1)^r \underbrace{ \left( \frac{ \cos[(m-r)\pi] - 1}{2i(m-r)} - \frac{\cos[(m-r)\pi] - 1}{2i(m-r)} \right) }_{ =0 } + \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{1}{2^{2m}} \binom{2m}{m} \frac{\pi}{2} \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}$$

 

(ii) $n=2m+1$ (odd) $$\begin{align} J_{2m+1} &= \int_0^{ \frac{ \pi }{ 2 } } \sin^{2m+1} x \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \left( \frac{ \textrm{e}^{ix} - \textrm{e}^{-ix} }{ 2i } \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left( \textrm{e}^{ix} - \textrm{e}^{-ix} \right)^{2m+1} \, \textrm{d}x \\ &= \int_0^{ \frac{ \pi }{ 2 } } \frac1{(2i)^{2m+1}} \left[ \sum_{r=0}^{2m+1} \binom{2m+1}{r} \textrm{e}^{i(2m+1-r)x} (-1)^r \textrm{e}^{-irx} \right] \, \textrm{d}x \qquad \textrm{by binomial theorem} \\ &= \frac1{(2i)^{2m+1}} \sum_{r=0}^{2m+1} \binom{2m+1}{r} (-1)^r \underbrace{ \left( \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \right) }_{ I_r } \\ \end{align}$$ We note that, $$\begin{align} I_r &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= \frac1{i(2m+1-2r)} \left( \textrm{e}^{i(m-r)\pi + \frac{i\pi}{2} } - 1 \right) \\ &= \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ \\ I_{2m+1-r} &= \int_0^{ \frac{ \pi }{ 2 } } \textrm{e}^{-i(2m+1-2r)x} \, \textrm{d}x \\ &= \left[ - \frac1{i(2m+1-2r)} \textrm{e}^{i(2m+1-2r)x} \right]_0^{ \frac{ \pi }{ 2 } } \\ &= - \frac1{i(2m+1-2r)} \left( \textrm{e}^{-i(m-r)\pi - \frac{i\pi}{2} } - 1 \right) \\ &= - \frac{ - i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } \\ &= \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \\ \end{align}$$ As with the even case above, we separate the summation into two parts, $$\begin{align} \sum_{r=0}^{2m+1} = \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \end{align}$$ $$\begin{align} \Rightarrow \qquad J_{2m+1} &= \frac1{(2i)^{2m+1}} \left( \sum_{r=0}^{m} + \sum_{r=m+1}^{2m+1} \right) \binom{2m+1}{r} (-1)^r I_r \\ \end{align}$$ For the second summation, let $k = 2m+1-r$ and it gives $$\begin{align} \Rightarrow \qquad J_{2m+1} &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{2m+1-k} (-1)^{2m+1-k} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \left[ \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r I_r + \sum_{k=0}^{m} \binom{2m+1}{k} (-1)^{k-1} I_{2m+1-k} \right] \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \Big( I_r - I_{2m+1-r} \Big) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^r \left( \frac{ i \cos [(m-r)\pi] - 1 }{ i(2m+1-2r) } - \frac{ i \cos [(m-r)\pi] + 1 }{ i(2m+1-2r) } \right) \\ &= \frac{1}{(2i)^{2m+1}} \sum_{r=0}^{m} \binom{2m+1}{r} (-1)^{r+1} \left( \frac{ 2 }{ i(2m+1-2r) } \right) \\ &= \frac{(-1)^m}{2^{2m}} \sum_{r=0}^{m} \binom{2m+1}{r} \frac{ (-1)^{r} }{ (2m+1-2r) } \qquad \textrm{(Dirichlet beta function?)}\\ &= \frac{(-1)^m}{2^{2m}} \left( \frac{ 2^{2m+1} \Gamma \left( \frac12 - m \right) \Gamma(m+1) }{ \sqrt{\pi} (2m+1) } \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \Gamma \left( \frac12 - m \right) \\ &= \frac{ 2(-1)^m m! }{ \sqrt{\pi} (2m+1) } \left[ (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \right] \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align}$$ since $$\begin{align} \Gamma \left( -m + \frac12 \right) &= \frac{ \Gamma \left( -m +1 + \frac12 \right) }{ \left( -m + \frac12 \right) } \\ &= \frac{ \Gamma \left( -m + m + \frac12 \right) }{ \left( -m + \frac12 \right) \left( -m + 1 + \frac12 \right) \left( -m + 2 + \frac12 \right) \cdots \left( -m + (m - 1) + \frac12 \right) } \\ &= \frac{ \Gamma \left( \frac12 \right) }{ (-1)^m \frac{ (2m-1)(2m-3)(2m-5) \cdots 3 \cdot 1 }{ 2^m } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)(2m-1)(2m-2)(2m-3)(2m-4) \cdots 1 }{ 2^m (2m)(2m-2)(2m-4) \cdots 2 } } \\ &= \frac{ \sqrt{\pi} }{ (-1)^m \frac{ (2m)! }{ 2^{2m} m! } } \\ &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi } \end{align}$$

 

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Method 3. Integration by substitution and the Euler beta and gamma functions:

Use the substitution $$\begin{align} u &= \sin x \\ \textrm{d}u &= \cos x \, \textrm{d}x = \sqrt{ 1 - \sin^2x }\, \textrm{d}x = \sqrt{ 1 - u^2 }\, \textrm{d}x \end{align}$$ and the integral reads $$\begin{align} \Rightarrow \qquad J_n &= \int_0^{\frac{\pi}{2}} \sin^nx \, \textrm{d}x = \int_0^1 u^n \left( 1 - u^2 \right)^{-\frac12} \, \textrm{d}u \end{align}$$ We make the substitute: $$\begin{align} t &= u^2 \\ \textrm{d}t &= 2u \, \textrm{d}u = 2t^{\frac12} \, \textrm{d}u \end{align}$$ which gives $$\begin{align} \Rightarrow \qquad J_n &= \int_0^1 t^{ \frac{n}{2} } \left( 1 - t \right)^{-\frac12} \, \frac{ \textrm{d}t }{ 2t^{\frac12} } \\ &= \frac12 \int_0^1 t^{ \frac{n}{2} - \frac12 } \left( 1 - t \right)^{-\frac12} \, \textrm{d}t \\ &= \frac12 B \left( \frac{n+1}{2}, \frac12 \right) \\ &= \frac12 \frac{ \Gamma\left( \frac{n}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{n}{2}+1 \right) } \end{align}$$ where the Euler beta and gamma functions are defined by $$\begin{align} B(p,q) &= \int_0^1 t^{p-1} (1-t)^{q-1} \, \textrm{d}t \\ \Gamma(p) &= \int_0^\infty t^{p-1}\textrm{e}^{-t} \, \textrm{d}t \end{align}$$ and $$\begin{align} B(p,q) = \frac{ \Gamma(p) \Gamma(q) }{ \Gamma(p+q) } \qquad \textrm{and} \qquad \Gamma(p+1) = p \Gamma(p) \end{align}$$ Noting that $\Gamma \left( \frac12 \right) = \sqrt{\pi}$ (see below), we find, for $m \in \mathbb N_0$, $$\begin{align} \Gamma \left( m + \frac12 \right) &= \left( m - \frac12 \right) \Gamma \left( m - \frac12 \right) \\ &= \underbrace{ \left( m - \frac12 \right) \left( m - \frac32 \right) \cdots \left( \frac12 \right) }_{n\;\textrm{terms}} \Gamma \left( \frac12 \right) \\ &= \frac{ (2m-1) (2m-3) \cdots 3 \cdot 1 }{ 2^m } \Gamma \left( \frac12 \right) \\ &= \frac{ (2m) (2m-1) (2m-2) (2m-3) (2m-4) \cdots 3 \cdot 2 \cdot 1 }{ 2^m (2m) (2m-2) (2m-4) \cdots 4 \cdot 2 } \underbrace{ \Gamma \left( \frac12 \right) }_{=\sqrt{\pi}} \\ &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \end{align}$$ where we have used $$\begin{align} \left[ \Gamma \left( \frac12 \right) \right]^2 &= B\left( \frac12, \frac12 \right) \underbrace{ \Gamma(1) }_{0!=1} \\ &= \int_0^1 t^{-\frac12} (1-t)^{-\frac12} \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ t(1-t)} } \, \textrm{d}t \\ &= \int_0^1 \frac1{ \sqrt{ \frac14 - \left( t - \frac12 \right)^2 }} \, \textrm{d}t \end{align}$$ and by the substitution $$\begin{align} t-\frac12 &= \frac12\sin v \\ \textrm{d}t &= \frac12\cos v\,\textrm{d}v \\ \end{align}$$ which gives $$\begin{align} &\Rightarrow& \left[ \Gamma \left( \frac12 \right) \right]^2 &= \int_0^{\frac{\pi}{2}} \frac1{ \sqrt{ \frac14 \left( 1 - \sin^2v \right) }} \, \frac12\cos v\,\textrm{d}v = \pi \\ &\Rightarrow& \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \end{align}$$

 

(i) $n=2m$ (even) $$\begin{align} J_{2m}&= \frac12 \frac{ \Gamma\left( \frac{2m}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( m+1 \right) } \\ &= \frac12 \frac{ \left( \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right) \sqrt{\pi} }{ m! } \\ &= \frac{ (2m)! }{ 2^{2m} (m!)^2 } \frac{\pi}{2} \qquad \square \end{align}$$

 

(ii) $n=2m+1$ (odd) $$\begin{align} J_{2m+1}&= \frac12 \frac{ \Gamma\left( \frac{2m+1}{2} + \frac12 \right) \Gamma\left( \frac12 \right) }{ \Gamma\left( \frac{2m+1}{2}+1 \right) } \\ &= \frac12 \frac{ \Gamma\left( m + 1 \right) \Gamma\left( \frac12 \right) }{ \Gamma \left( m + \frac32 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \Gamma \left( m + \frac12 \right) } \\ &= \frac12 \frac{ m! \sqrt{\pi} }{ \left( m + \frac12 \right) \left[ \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi} \right] } \\ &= \frac{ 2^{2m+1} (m!)^2 }{ (2m+1)! } \qquad \square \end{align}$$

 

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A summary:

 

$$\begin{align} \int_0^{\frac{\pi}{2}}\sin^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\sin^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m}x\,\textrm{d}x=\frac{(2m)!}{2^{2m}(m!)^2}\frac{\pi}{2} \\ \\ \int_0^{\frac{\pi}{2}}\cos^{2m+1}x\,\textrm{d}x=\frac{2^{2m}(m!)^2}{(2m+1)!} \end{align}$$

and $$\begin{align} \Gamma \left( \frac12 \right) &= \sqrt{ \pi } \\ \Gamma \left( m + \frac12 \right) &= \frac{ (2m)! }{ 2^{2m} m! } \sqrt{\pi}, \quad m \in \mathbb Z_0^+ \\ \Gamma \left( -m + \frac12 \right) &= (-1)^m \frac{ 2^{2m} m! }{ (2m)! } \sqrt{ \pi }, \quad m \in \mathbb Z_0^+ \end{align}$$